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61. 62. 63. 64. X f(x) X f(x) X f(x) X 式 f(x) -1 0 1 2 0 224 -10 13 0224 -10 23 0224 1 2 3 224 -1 0

61. 62. 63. 64. X f(x) X f(x) X f(x) X 式 f(x) -1 0 1 2 0 224 -10 13 0224 -10 23 0224 1 2 3 224 -1 0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Parts 61-64 please… please use the second image as reference for the work that needs to be done
**Problem IV:** Find a linear polynomial \( f(x) \) which is the best least squares fit to the following data. **Data Sets:** 1. **Set 61:** - \( x: -1, 0, 1, 2 \) - \( f(x): 0, 2, 2, 4 \) 2. **Set 62:** - \( x: -1, 0, 1, 3 \) - \( f(x): 0, 2, 2, 4 \) 3. **Set 63:** - \( x: -1, 0, 2, 3 \) - \( f(x): 0, 2, 2, 4 \) 4. **Set 64:** - \( x: -1, 1, 2, 3 \) - \( f(x): 0, 2, 2, 4 \) 5. **Set 65:** - \( x: -2, 0, 1, 2 \) - \( f(x): 0, 2, 2, 4 \) 6. **Set 66:** - \( x: -2, -1, 0, 2 \) - \( f(x): 0, 2, 2, 4 \) 7. **Set 67:** - \( x: -1, 0, 1, 2 \) - \( f(x): 1, 2, 2, 3 \) 8. **Set 68:** - \( x: -1, 0, 1, 3 \) - \( f(x): 1, 2, 2, 3 \) 9. **Set 69:** - \( x: -1, 0, 2, 3 \) - \( f(x): 1, 2, 2, 3 \) 10. **Set 70:** - \( x: -1, 1, 2, 3 \) - \( f(x): 1, 2, 2, 3 \) 11. **Set 71:** - \( x: -1, 0,
Transcribed Image Text:**Problem IV:** Find a linear polynomial \( f(x) \) which is the best least squares fit to the following data. **Data Sets:** 1. **Set 61:** - \( x: -1, 0, 1, 2 \) - \( f(x): 0, 2, 2, 4 \) 2. **Set 62:** - \( x: -1, 0, 1, 3 \) - \( f(x): 0, 2, 2, 4 \) 3. **Set 63:** - \( x: -1, 0, 2, 3 \) - \( f(x): 0, 2, 2, 4 \) 4. **Set 64:** - \( x: -1, 1, 2, 3 \) - \( f(x): 0, 2, 2, 4 \) 5. **Set 65:** - \( x: -2, 0, 1, 2 \) - \( f(x): 0, 2, 2, 4 \) 6. **Set 66:** - \( x: -2, -1, 0, 2 \) - \( f(x): 0, 2, 2, 4 \) 7. **Set 67:** - \( x: -1, 0, 1, 2 \) - \( f(x): 1, 2, 2, 3 \) 8. **Set 68:** - \( x: -1, 0, 1, 3 \) - \( f(x): 1, 2, 2, 3 \) 9. **Set 69:** - \( x: -1, 0, 2, 3 \) - \( f(x): 1, 2, 2, 3 \) 10. **Set 70:** - \( x: -1, 1, 2, 3 \) - \( f(x): 1, 2, 2, 3 \) 11. **Set 71:** - \( x: -1, 0,
**Problem 6.** Find a linear polynomial which is the best least squares fit to the following data: \[ \begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -3 & -2 & 1 & 2 & 5 \\ \end{array} \] We are looking for a function \( f(x) = c_1 + c_2 x \), where \( c_1, c_2 \) are unknown coefficients. The data of the problem give rise to an overdetermined system of linear equations in variables \( c_1 \) and \( c_2 \): \[ \begin{cases} c_1 - 2c_2 = -3, \\ c_1 - c_2 = -2, \\ c_1 = 1, \\ c_1 + c_2 = 2, \\ c_1 + 2c_2 = 5. \end{cases} \] This system is inconsistent. We can represent the system as a matrix equation \( Ac = y \), where: \[ A = \begin{pmatrix} 1 & -2 \\ 1 & -1 \\ 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad c = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}, \quad y = \begin{pmatrix} -3 \\ -2 \\ 1 \\ 2 \\ 5 \end{pmatrix}. \] The least squares solution \( c \) of the above system is a solution of the normal system \( A^T Ac = A^T y \): \[ \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ -2 & -1 & 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 1 & -1 \\ 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}
Transcribed Image Text:**Problem 6.** Find a linear polynomial which is the best least squares fit to the following data: \[ \begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -3 & -2 & 1 & 2 & 5 \\ \end{array} \] We are looking for a function \( f(x) = c_1 + c_2 x \), where \( c_1, c_2 \) are unknown coefficients. The data of the problem give rise to an overdetermined system of linear equations in variables \( c_1 \) and \( c_2 \): \[ \begin{cases} c_1 - 2c_2 = -3, \\ c_1 - c_2 = -2, \\ c_1 = 1, \\ c_1 + c_2 = 2, \\ c_1 + 2c_2 = 5. \end{cases} \] This system is inconsistent. We can represent the system as a matrix equation \( Ac = y \), where: \[ A = \begin{pmatrix} 1 & -2 \\ 1 & -1 \\ 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad c = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}, \quad y = \begin{pmatrix} -3 \\ -2 \\ 1 \\ 2 \\ 5 \end{pmatrix}. \] The least squares solution \( c \) of the above system is a solution of the normal system \( A^T Ac = A^T y \): \[ \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ -2 & -1 & 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 1 & -1 \\ 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}
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